Real questions rarely hand you F_net. They hand you a scene: a 5.0 kg sled, a rope towing it with 30 N of tension, 10 N of friction from the snow.
Finding the sled's acceleration takes three moves, and you own all three.
Move one: draw the free-body diagram, every force on the object as a labeled arrow from its center.
For the sled, that is four arrows: the tension forward, the friction backward, the weight downward, and the normal force upward.
One direction you supply yourself. Friction opposes the sliding, and the towed sled slides forward, so the friction on it acts backward. When a question gives a friction force without naming its direction, you point it against the sliding yourself.
Move two: find the net force, treating each line separately. The vertical forces balance here, so the net force is the horizontal difference: F_net = 30 N − 10 N = 20 N, forward.
Move three: apply a = F_net/m with the usual solving steps.
The acceleration points with the net force: forward, the tension's way.
A 5.0 kg sled is towed with 30 N of tension while 10 N of friction acts on it. Calculate its acceleration.
Write down the values in the question:
tension = 30 N forward
friction = 10 N backward
m = 5.0 kg
a = ?
Find the net force from the diagram:
F_net = 30 N − 10 N
F_net = 20 N forward
Write down the equation:
a = F_net/m
Substitute in the values:
a = 20 ÷ 5.0
Simplify:
a = 4.0 m/s²
The sled's acceleration is 4.0 m/s², in the direction of the tension.
A rope tows a 12 kg crate over a level floor with 50 N of tension, while 14 N of friction acts on the crate. Enter the crate's acceleration in m/s².
A rope tows a 16 kg crate with 96 N of tension while 24 N of friction acts on it. Calculate the crate's acceleration.
Write down the values in the question:
tension = 96 N forward
friction = 24 N backward
m = 16 kg
a = ?
Find the net force from the diagram:
F_net = 96 N − 24 N
F_net = 72 N forward
Write down the equation:
a = F_net/m
Substitute in the values:
a = 72 ÷ 16
Simplify:
a = 4.5 m/s²
The crate's acceleration is 4.5 m/s², in the direction of the tension.
A dock worker tows a 20 kg cart with a strap at 84 N of tension, while 30 N of friction acts on the cart. Enter the cart's acceleration in m/s².
Some scenes bring more than two horizontal forces. Forces in the same direction add; a force in the opposite direction subtracts. Two ropes in the same direction add their tensions first, and the friction then comes off the total.
Two ropes tow a 50 kg ice sled, with tensions of 30 N and 24 N in the same direction, while 14 N of friction acts on the sled. Enter the sled's acceleration in m/s².
A rope tows a 30 kg garden cart with 120 N of tension, while 45 N of friction acts on the cart. Enter the cart's acceleration in m/s².
A hauler's rope tows a wagon loaded with firewood, total mass 75 kg, with 150 N of tension, while friction from the ground acts on the wagon at 30 N. Enter the wagon's acceleration in m/s².
A 36 kg sled is towed with 95 N of tension, while 23 N of friction acts on it. Enter the sled's acceleration in m/s².